BHU PMT BHU PMT (Mains) Solved Paper-2009

  • question_answer
    In a hall, a person receives direct sound waves from a source 120 m away. He also receives waves from the same source which reach him after being reflected from the 25 m high ceiling at a point half-way between them. The two waves interfere constructively for wavelengths (in metre) of

    A)  \[10,5,\frac{5}{2}\]                        

    B)  \[20,\frac{20}{3},\frac{20}{5}.....\]

    C)  \[30,20,10.....\]               

    D)  \[35,25,15.....\]

    Correct Answer: B

    Solution :

                     Path difference \[\Delta x(SA+AP)-SP\]                 \[\Rightarrow \] \[\Delta x=(65+65)-120\]                 \[\Rightarrow \]               \[\Delta x=10\,m\] But at A the wave suffers reflection at the surface of rigid/fixed end or denser medium hence the wave must suffer an additional path change of \[\frac{\lambda }{2}\]on a phase change of\[\pi \].                                 \[\Rightarrow \]Net path difference                 \[=\left( 10-\frac{\lambda }{2} \right)\]                 For maxima (constrictive interference) Net path difference                 \[=(2n)\frac{\lambda }{2},n=0,1,2,3....\]                 \[10-\frac{\lambda }{2}=(2n)\frac{\lambda }{2};n=0,1,2,....\]                 \[\Rightarrow \]               \[10=(2n+1)\frac{\lambda }{2};n=0,1,2,.....\]                 \[\Rightarrow \]               \[\lambda =\frac{20}{2n+1};n=0,1,2,....\]                 \[\Rightarrow \]               \[\lambda =20,\frac{20}{3},\frac{20}{5},\frac{20}{7}.....\]


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